Partial Fraction Decomposition Question? 10 Points

Find the partial fraction decomposition of the rational function.

3x-9/x^3+3x

**An explanation and the answer would be sweet. Thanks.

Here's the general principle: Any rational fraction P(x) / Q(x) with the order of P(x) smaller than Q(x) can be rewritten as the sum of a bunch of fractions that have the factors if Q(x) in the denominator, and polynomials with degree LESS than those factors in the numerator. Uh, "HUH?", I hear you say... well let's use your problem as an example:

P(x) = 3x-9
Q(x) = x^3 + 3x = x(x^2 + 3)

So we can rewrite (3x-9) / (x^3 + 3x) as:

(Ax + B)/(x^2 + 3) + C/x

Now we just need to set these equal and solve for A, B, and C:

(Ax + B)/(x^2 + 3) + C/x = (3x-9) / (x^3 + 3x) {multiply both sides by x^3 + 3x}
(Ax + B) * x + C * (x^2 + 3) = 3x-9
Ax^2 + Bx + Cx^2 + 3C = 3x-9
(A+C)x^2 + Bx + 3C = 3x - 9

Now, the values for EACH power of x have to be equal on both sides! So we get 3 equations:

A+C = 0
B = 3
3C = -9

We can quickly solve these as B=3, C=-3, and A=3. So the result is:

(3x-9) / (x^3 + 3x) = (3x + 3)/(x^2 + 3) - 3/x

And that's the answer!

Happy to help. There are two tricks I didn't mention. 1) If P(x) has equal or greater degree than Q(x), just divide first. You'll get a polynomial + a fractional remainder.

2) If the fraction is like 2x-1/(x+1)^2, then you can decompose it into A/(x+1) + B/(x+1)^2.

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3x – 9 ......... A ........ Bx + C
-------- --- = -------- + ------------
x(x² + 3) ...... x ......... x² + 3


Multiply through by x(x²+3) to clear fractions; then collect like terms and compare coeffficients.

3x – 9 = A(x² + 3) + x(Bx + C)

3x – 9 = Ax² + 3A + Bx² + Cx

0x² + 3x – 9 = (A + B)x² + Cx + 3A

Clearly:

A + B = 0
C = 3
A= –3 which tells you that B = +3. So substituting:

3x – 9 ......... –3 ........ 3x + 3
-------- --- = -------- + ------------
x(x² + 3) ...... x ......... x² + 3

Which you can rearrange, factor parts,whatever you like.

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